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### Transcript of Hamiltonian Equations

• MechanicsPhysics 151

Lecture 18Hamiltonian Equations of Motion

(Chapter 8)

We are starting Hamiltonian formalismHamiltonian equation Today and 11/26Canonical transformation 12/3, 12/5, 12/10Close link to non-relativistic QM

May not cover Hamilton-Jacobi theoryCute but not very relevant

What shall we do in the last 2 lectures?Classical chaos?Perturbation theory?Classical field theory?Send me e-mail if you have preference!

• Hamiltonian Formalism

Newtonian Lagrangian HamiltonianDescribe same physics and produce same resultsDifference is in the viewpoints

Symmetries and invariance more apparentFlexibility of coordinate transformation

Hamiltonian formalism linked to the development ofHamilton-Jacobi theoryClassical perturbation theoryQuantum mechanicsStatistical mechanics

• Lagrange Hamilton

Lagranges equations for n coordinates

n equations 2n initial conditions

Can we do with 1st-order differential equations?Yes, but youll need 2n equationsWe keep and replace with something similar

We take the conjugate momenta

0i i

d L Ldt q q

=

1, ,i n= 2nd-order differential

equation of n variables

( 0)iq t = ( 0)iq t =

iq iq( , , )j j

ii

L q q tp

q

• Configuration Space

We considered as a point in an n-dim. spaceCalled configuration spaceMotion of the system A curve in the config space

When we take variations,we consider and asindependent variables

i.e., we have 2n independent variables in n-dim. spaceIsnt it more natural to consider the motion in 2n-dim space?

1( , , )nq q

( )i iq q t=

iq iq

• Phase Space

Consider coordinates and momenta as independentState of the system is given byConsider it a point in the 2n-dimensional phase space

We are switching theindependent variables

A bit of mathematical trickis needed to do this

1 1( , , , , , )n nq q p p

( )( )

i i

i i

q q tp p t

=

=( , , ) ( , , )i i i iq q t q p t

• If and

Legendre Transformation

Start from a function of two variablesTotal derivative is

Define and consider its total derivative

i.e. g is a function of u and y

( , )f x y

g f ux

f fdf dx dy udx vdyx y

= + +

( )dg df d ux udx vdy udx xdu vdy xdu= = + =

g vy

=

g xu

=

f L= ( , ) ( , )x y q q=

( , ) ( , )L q q g p q L pq =

This is what we need

• Hamiltonian

Define Hamiltonian:Total derivative is

Lagranges equations say

This must be equivalent to

( , , ) ( , , )i iH q p t q p L q q t=

i i i i i ii i

L L LdH p dq q dp dq dq dtq q t

= +

Opposite sign from Legendre transf.

ii

L pq

=

i i i iLdH q dp p dq dtt

=

i ii i

H H HdH dp dq dtp q t

= + +

Putting them together gives

• Hamiltons Equations

We find and

2n equations replacing the n Lagranges equations1st-order differential instead of 2nd-orderSymmetry between q and p is apparent

There is nothing new We just rearranged equationsFirst equation links momentum to velocity

This relation is given in Newtonian formalismSecond equation is equivalent to Newtons/Lagranges equations of motion

ii

H qp

=

ii

H pq

=

H Lt t

=

• Particle under Hookes law force F = kx

Hamiltons equations

Quick Example

2 2

2 2m kL x x=

2 2

22

2 2

2 2

m kH xp L x x

p k xm

= = +

= +

Lp mxx

= =

Hp kxx

= =

H pxp m

= =

Replace withx pm

Usual harmonic oscillator

• Energy Function

Definition of Hamiltonian is identical to the energy

function

Difference is subtle: H is a function of (q, p, t)

This equals to the total energy ifLagrangian isConstraints are time-independent

This makesForces are conservative

This makes

( , , ) ( , , )ii

Lh q q t q L q q tq

=

0 1 2( , ) ( , ) ( , )i j kL L q t L q t q L q t q q= + +

2 ( , ) j kT L q t q q=

0 ( )V L q=

See Lecture 4, or Goldstein Section 2.7

• Hamiltonian and Total Energy

If the conditions make h to be total energy, we can skip calculating L and go directly to H

For the particle under Hookes law force

This works often, but not alwayswhen the coordinate system is time-dependent

e.g., rotating (non-inertial) coordinate systemwhen the potential is velocity-dependent

e.g., particle in an EM field

22

2 2p kH E T V xm

= = + = +

Lets look at this

• Particle in EM Field

For a particle in an EM field

Wed be done if we were calculating h

For H, we must rewrite it using

2

2 i i imL x q qA x= + We cant jump on H = E

because of the last term, but

i i ip mx qA= +2

( )2

ii i i

mxH mx qA x L q= + = + This is in fact E

2( )( , )2

i ii i

p qAH x p qm

= +

i i ip mx qA= +

• Particle in EM Field

Hamiltons equations are

Are they equivalent to the usual Lorentz force?Check this by eliminating pi

2( )( , )2

i ii i

p qAH x p qm

= +

i ii

i

p qAHxp m

= =

j j j

ii i i

p qA AHp q qx m x x

= =

( ) ji i ii i

Ad mx qA qx qdt x x

+ =

( ) ( )i i id mv qE qdt

= + v BA bit of work

• Conservation of Hamiltonian

Consider time-derivative of Hamiltonian

H may or may not be total energyIf it is, this means energy conservationEven if it isnt, H is still a constant of motion

( , , )dH q p t H H Hq pdt q p t

Hpq qpt

= + +

= + +

Hamiltonian is conserved if it does not depend explicitly on t

• Cyclic Coordinates

A cyclic coordinate does not appear in LBy construction, it does not appear in H either

Hamiltons equation says

Exactly the same as in the Lagrangian formalism

(H q , , ) (i ip t q p qL= , , )q t

0Hpq

= =

Conjugate momentum of a

cyclic coordinate is conserved

• Cyclic Example

Central force problem in 2 dimensions

Cyclic variable drops off by itself

2 2 2( ) ( )2mL r r V r= +

rp mr=2p mr =

22

2

22

2

1 ( )2

1 ( )2

r

r

pH p V rm r

lp V rm r

= + +

= + +

is cyclic constp l = =Hamiltons equations

rprm

=2

3

( )r

l V rpmr r

=

• Going Relativistic

Practical approachFind a Hamiltonian that works

Does it represent the total energy?

Purist approachConstruct covariant Hamiltonian formalism

For one particle in an EM field

Dont expect miraclesFundamental difficulties remain the same

• Practical Approach

Start from the relativistic Lagrangian that works

It does equal to the total energyHamiltons equations

2 21 ( )L mc V= x

21i

ii

mvLpv

= =

Did this last time

2 2 2 4 ( )H h p c m c V= = + + x

2

2 2 2 4i i

ii

p c pHxp mp c m c

= = =

+i i

i i

H Vp Fx x

= = =

• Practical Approach w/ EM Field

Consider a particle in an EM field

Hamiltonian is still total energy

Difference is in the momentum

2

2 2 2 2 2 4

H m c q

m v c m c q

= +

= + +

Can be easily checked

2 21 ( ) ( )L mc q q = + x v A

i i ip m v qA= +2 2 2 4( )H q c m c q= + +p A

Not the usual linear momentum!

• Practical Approach w/ EM Field

Consider H q

It means that is a 4-vector,and so is

This particular Hamiltonian + canonical momentum transforms as a 4-vector

True only for well-defined 4-potential such as EM field

2 2 2 4( )H q c m c q= + +p A

2 2 2 2 4( ) ( )H q q c m c =p A constant

( , )H q c q c p A( , )H cp Similar to 4-momentum (E/c, p) of

a relativistic particle

Remember p here is not the linear momentum!

• Purist Approach

Covariant Lagrangian for a free particle

We know that p0 is E/cWe also know that x0 is ct

Generally true for any covariant LagrangianYou know the corresponding relationship in QM

12 mu u

=

p muu

= =

2p p

Hm

=

Energy is the conjugate momentum of time

• Purist Approach

Value of Hamiltonian is

What is important is Hs dependence on p

Hamiltons equations

Time components are

dx H pd p m

= =

0dp Hd x

= =

2

2 2p p mcH

m

= =

4-momentum conservation

This is constant!

( )d ct E cd mc

= =( ) 0d E cd

= Energy definitionand conservation

• Purist Approach w/ EM Field

With EM field, Lagrangian becomes

Hamiltons equations are

A bit of work can turn them into

12( , )x u mu u qu A

= + p mu qA

= +

( )( )2 2

mu u p qA p qAH

m

= =

dx H p qAd p m

= =

= =

du A Am q u Kd x x

= = 4-force

• EM Field and Hamiltonian

In Hamiltonian formalism, EM field always modify the canonical momentum as

A handy rule:

Often used in relativistic QM to introduce EM interaction

0Ap p qA = +